A = .3*g = 2.94 m/s²
<span>t = v/a = 9/2.94 = 3.061 sec </span>
<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:

Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)


We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2

b) The acceleration of point A is:

(aA/D)t = ADαj

The acceleration of point E is:
(aE/D)t = -EDαj

Answer:
The correct answer is Gamma Rays.
Explanation:
Answer:
18.375
Explanation:
Mean = total of measured values / no of elements of sample
Total = .15 + .89 +1.11+1.46 +2.78 + 3.12 + 4.30 + 4.59 + 4.92 + 6.42 + 7.20+8.04+8.21+12.13+31.86+32.53+33.82+36.60+ 72.99
= 273.03
Mean = 273 .03 / 19
= 14.37
( .15 - 14.37 )² +( 0.89 - 14.37 )²+( 1.11- 14.37 )²+( 1.46 - 14.37 )²+( 2.78 - 14.37 )²+( 3.12 - 14.37 )²+( 4.30 - 14.37 )²+( 4.59 - 14.37 )²+( 4.92 - 14.37 )²+( 6.42 - 14.37 )²+( 7.20 - 14.37 )²+( 8.04 - 14.37 )²+( 8.21 - 14.37 )²+( 12.13 - 14.37 )²+( 31.86 - 14.37 )²+( 32.53 - 14.37 )²+( 33.82 - 14.37 )²+( 36.60 - 14.37 )²+( 72.99 - 14.37 )²
= 202.20 +181.71+ 175.82 + 166.66+ 134.32 + 126.56 + 101.40 +95.64 + 89.30 +63.20 +51.40 +40.06 +37.94+5.01 +305.90 +329.78 +378.30 +494.17 +3436.30
=6415.67
Standard deviation =

= 18.375
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